Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Mathematics authors agree that $+,-,/,\times$ are basic operators. There are also logical operators like $\text{or, and, xor}$ and the unary negation operator $\neg$.

Where there seems to be a disagreement, however, is whether certain symbols used in the composition of propositions qualify as operators. By this definition:

an operation is an action or procedure which produces a new value from one or more input values, called "operands"

symbols that denote a relation should also be operators.

For instance, $a<b$ takes integral operands and returns a value in the Boolean domain. Even something like $x \in S$ should be an operator that takes an element and a set, and also returns a truth value. The same could be said for the equality operator $=$.

Computer scientists call these 'relational operators', but mathematicians rarely do this. Is there a reason for this discrimination?

share|improve this question
1  
Operator theory is a whole other thing. –  Asaf Karagila Feb 20 '13 at 20:14
    
Operator and operation may be used differently in English –  GEdgar Feb 20 '13 at 20:37
    
Why do you say that $\in$ should be an operator that takes an element and a set, and returns a truth value? –  Trevor Wilson Feb 20 '13 at 22:05
1  
@TrevorWilson, perhals could should be the word... I am trying to reason this to make sense of how the symbol can be used as a predicate in the definition of a set (please see comment to Hagen von Eitzen below) –  jesterII Feb 20 '13 at 22:09

5 Answers 5

up vote 2 down vote accepted

I believe that the term "operator" in mathematics usually refers a function $S^n\to S$ for some domain $S$. For example, the $+$ operator for real numbers is a function $\mathbb R^2\to\mathbb R$. The "binary operation" of a group is a mapping $G^2\to G$.

In contrast, a "relation" such as $<$ on real numbers is not thought of as a function, but as a subset, in this case a subset of $\mathbb R^2$. We could think of it as a function from $\mathbb R^2$ to a special two-element set $\mathbb B = \{\bf\text{true}, \bf\text{false}\}$. This idea is isomorphic to the standard notion, since a subset $R\subset \mathbb R^2$ can be identified with the function which takes an argument $x$ to true if $x\in R$ and to false if not. The two definitions are isomorphic, but the jargon is different.

Now consider the $\lor$ operator, which is a mapping from $\mathbb B^2\to\mathbb B$, and which usually is called an operator. Despite the fact that it is a mapping with a codomain of $\mathbb B$, it is not normally thought of as a relation. As far as I know nobody ever thinks of the $\lor$ operator as the relation $\{ \langle\bf\text{false}, \bf\text{true}\rangle, \langle\bf\text{true}, \bf\text{false}\rangle, \langle\bf\text{true}, \bf\text{true}\rangle \}$.

You asked about the $\in$ relation, and it is often considered just that, a relation, although for technical reasons $\in$ in its fullest generality is not actually a relation. (Relations are sets, and $\in$ is too big to be a set.) But in limited domains one can consider the subset of $X\times \mathcal P(X)$ consisting of all pairs $\langle x, Y\rangle$ such that $x\in Y$, and this can be considered a definition of a limited version of $\in$. I have never seen $\in$ considered to be an operator.

In the implementation of computer languages there is no reason to make this distinction, and all such mappings are considered operators, and are implemented the same way: typically the argument values are taken off a stack and replaced with the result value. The types of the argument and result values don't matter here, so computer programmers and language implementors don't make the distinction.

share|improve this answer
    
Great response, and so what would be the terminology reserved for these symbols? –  jesterII Feb 20 '13 at 19:58
    
For which symbols? –  MJD Feb 20 '13 at 20:05
1  
$\vee$ is often thought of as a "logical connective" rather than an "operator". –  Hurkyl Feb 20 '13 at 20:05
    
@Hurkyl That is a good point. It can be a connector in formulas, and there is a related but separate matter of the $\lor$ operator that applies to the values of the formulas. I am only discussing the latter one here. –  MJD Feb 20 '13 at 20:06

A statement like $(x\in S)\in\{\text{true},\text{false}\}$ is not even a well-formed formula in the usual language of set theory.

share|improve this answer
1  
Why would that expression not be well-formed? –  jesterII Feb 20 '13 at 19:54
1  
Because $\in$ and $=$ can only stand between sets (in fact, in the absence of constants, only between variables) and thus produce an "atomic" logical expression (which can be combined using logical connectives and/or quantified) –  Hagen von Eitzen Feb 20 '13 at 20:04
    
There is an evident "transpose" to turn a logical expression with $n$ free variables into an $n$-ary relation or a $n$-ary function with codomain $\{ true, false \}$. It's not terribly unreasonable to transpose implicitly. –  Hurkyl Feb 20 '13 at 20:10
    
I am not sure about that, when I think of the definition of the set difference of $A$ and $B$, it is $A-B::=${$x \in A: x \notin B$} –  jesterII Feb 20 '13 at 21:27

Connotation. One uses the term "operator" for precisely those things one thinks of as "operating" on "operands".

A programmer is pretty much required to think of a relation as something that consumes its arguments and yields a value. But a mathematician is generally trained to use relations to form things like "statements" or "axioms" or "equations", rather than as a calculation.

share|improve this answer

The relational operators that you described are definitely operators. In the context, it probably wouldn't improve clarity to refer to them as such. An operator must act in a certain space, and having to declare that space and describe all its properties is probably to in the scope of the material. If you'd like to study the property of those relational operators, you'd study Boolean Algebra and its related subjects.

share|improve this answer

A serious disadvantage of treating logical connectives as operators returning a value (that is, as functions) is that in non-classical logics their "range" may fail to correspond to any two-element set $\{\top,\bot\}$, and according to the Wikipedia page Truth value, may even fail to correspond to any set whatsoever: "[n]ot all logical systems are truth-valuational in the sense that logical connectives may be interpreted as truth functions. For example, intuitionistic logic lacks a complete set of truth values...." (Note that in particular, the naive attempt to consider logical connectives as mapping into a three-element set $\{\top,\bot,\text{some third value}\}$ does not work.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.